The flow of traffic through a network of streets is show below in vehicles per hour. Solve the system for the traffic flow represented by xi, i = 1, 2, 3, 4, and 5. The i in xi is a subscript//
Model:
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Answer:
Rule: sum of inputs = sum of outputs
Basically whatever goes into a node, must come out (since no cars are stuck at any given intersection). There is an equal symmetric balance.
Take note of the arrow direction to determine if we have an input or output at any given node.
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Top left corner: The inputs are 400 (known quantity) and x2 (unknown value) leading to an output of x1 (also unknown).
This forms the equation
400+x2 = x1 which rearranges to x1 = x2+400
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Top right corner: We hve x1 and x3 leading to this node. They are both unknown. The outputs are x4 and 600, so we have this other equation
x1+x3 = 600+x4
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Bottom left corner: the only input is 300. The outputs are x2, x3 and x5
So we have
300 = x2+x3+x5
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Bottom right corner: The inputs are x4 and x5 leading to an output of 100
So,
x4+x5 = 100
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So far we have this system of equations
x1 = x2+400
x1+x3 = 600+x4
300 = x2+x3+x5
x4+x5 = 100
representing the four nodes.
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From here, I would get everything but the constants to one side turning the system mentioned above into this new system of equations
x1-x2 = 400
x1+x3-x4 = 600
x2+x3+x5 = 300
x4+x5 = 100
which is really this system
1x1-1x2+0x3+0x4+0x5 = 400
1x1+0x2+1x3-1x4+0x5 = 600
0x1+1x2+1x3+0x4+1x5 = 300
0x1+0x2+0x3+1x4+1x5 = 100
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Notice that last system forms the augmented matrix
which looks nasty, but it's not that bad really. Through a few row operations we can get to this matrix shown below
Note: I'm skipping showing the row operation steps because it simply takes too much space on a forum such as this; however, if you need me to go over that, then just let me know and I'll try to figure something out. You can use this calculator to do row operations and other linear algebra tasks.
The string of 0s in the last row indicate that we have
0x1+0x2+0x3+0x4+0x5 = 0
essentially the equation 0 = 0
The equation 0 = 0 is always true regardless of what x is, so we have infinitely many solutions. If we let x3 = s and x5 = t (s,t are integers such that
x4+x5 = 100
x4 = -x5+100
x4 = -t+100
based on what the second to last row is saying
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Then we have the second row
x2+x3+x5 = 300
solve for x2 to get
x2 = -x3+x5
and perform substitutions
x2 = -x3+x5
x2 = -s+t
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Finally, row 1 says
x1+x3+x5 = 700
isolate x1 and substitute:
x1+x3+x5 = 700
x1 = -x3-x5+700
x1 = -s-t+700
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So in summary:
x1 = -s-t+700
x2 = -s+t
x3 = s
x4 = -t+100
x5 = t
where again, s and t are integers such that
So as you can see, there isn't one fixed solution. Rather, there are a bunch of solutions. For instance, if (s,t) = (1,2), then
x1 = -s-t+700 = -1-2+700 = 697
x2 = -s+t = -1+2 = 1
x3 = s = 1
x4 = -t+100 = -2+100 = 98
x5 = t = 2
making one solution 5-tuple to be (x1,x2,x3,x4,x5) = (697,1,1,98,2)
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Another example:
Let (s,t) = (10,20)
x1 = -s-t+700 = -10-20+700 = 670
x2 = -s+t = -10+20 = 10
x3 = s = 10
x4 = -t+100 = -20+100 = 80
x5 = t = 20
So another solution is (x1,x2,x3,x4,x5) = (670, 10, 10, 80, 20)
There are a lot of ways to do this.